Question: $f\,^{\prime}(x)=-7e^x$ and $f(5)=24-7e^5$. $f(0) = $
Answer: Finding $f(x)$ We have $f'(x)=-7e^x$ and we want to find $f(x)$ : $\begin{aligned}f(x) &= \int f'(x)\,dx \\\\ & = \int (-7e^x)\,dx \\\\ & = {-7e^x} {+ C} \end{aligned}$ Finding $ C$ Goal: We need to find $ C$ such that $f(5)=24-7e^5$. Here's what we get when we plug in $5$ : $\begin{aligned}f(5)&={-7e^{5}} {+ C} \end{aligned}$ We are given that this must equal $24-7e^5$ : $24-7e^5 = {-7e^{5}} {+ C}$ Solving the equation gives us ${C=24}$. Finding $f(0)$ Now, we have that $f(x)={-7e^x} {+ 24}$. Let's find $f(0)$ by plugging in $0$ : $\begin{aligned}f(0)&=-7e^0 + 24\\\\ &=17 \end{aligned}$ The answer $f(0) = 17$